Move last element to front of a given Linked List

Move Last Element to Front of a Given Linked List


Problem statement: Given a Singly linked list, move the last element to the front of the linked list.

For example,
-> 1 -> 2 -> 3 -> 4 -> 5 -> NULL
-> 5 -> 1 -> 2 -> 3 -> 4 -> NULL

Here we traverse the linked list till the next pointer is NULL, store the last and second last element in two variables.

  NODE *secLast = NULL;
  NODE *last = head;
   while (last->next != NULL)
    {
        secLast = last;
        last = last->next;
    }

When we reach NULL pointer, assign the second last element’s next to NULL pointer.

    secLast->next = NULL;


Assign last element’s next to head.

    last->next = head;

At last assign the head to last element.

     head = last;

/*C code to move the last element to front of the Linked List */
#include <stdio.h>
#include <stdlib.h>
 
typedef struct node
{
    int data;
    struct node *next;
} NODE;
NODE *head = NULL;
 
 
NODE *newNode(int key)
{
    NODE *temp = (NODE *)malloc(sizeof(NODE));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
 
void moveToFront()
{
 
    if (head == NULL || (head)->next == NULL)
        return;
    NODE *secLast = NULL;
    NODE *last = head;
 
    while (last->next != NULL)
    {
        secLast = last;
        last = last->next;
    }
 
    secLast->next = NULL;
    last->next = head;
    head = last;
}
 
 
void printList()
{
    NODE *temp = head;
    if (temp == NULL)
    {
        printf("List is empty!\n");
        return;
    }
    printf("Linked List is\t");
    while (temp != NULL)
    {
        printf("%d--> ", temp->data);
        temp = temp->next;
    }
    printf("NULL\n");
}
 
 
int main()
{
    head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(3);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(5);
    head->next->next->next->next->next = NULL;
    printList();
    moveToFront();
    printf("Linked list after moving last element to front\n");
    printList();
    return 0;
}
$ ./a.out
Linked List is 1–> 2–> 3–> 4–> 5–> NULL
Linked list after moving last element to front
Linked List is 5–> 1–> 2–> 3–> 4–> NULL
$

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